Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
Sri Chaitanya IIT Academy.,India.
 A.P  T.S  KARNATAKA  TAMILNADU  MAHARASTRA  DELHI  RANCHI
A right Choice for the Real Aspirant
ICON Central Office - Madhapur - Hyderabad
SEC: Jr.Super60_STERLING BT
JEE-MAIN
Time: 09:00AM to 12:00PM
WTM-12
Date: 02-09-2023
Max. Marks: 300
KEY SHEET
PHYSICS
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2
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2
2
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1
3
2
1
2
6
CHEMISTRY
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36)
41)
46)
51)
56)
1
3
4
3
8
3
MATHEMATICS
61)
66)
71)
76)
81)
86)
2
1
4
4
6
7
SEC: Jr.Super60_STERLING BT
62)
67)
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2015
Page 1
Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
SOLUTIONS
PHYSICS
1.
For toppling
FL  mg 
2.
L
mg
F
2
2
  rFsin   constant
r1F1 sin 1  r2 F2 sin  2
F
O
3.
Mg
About O
F
3O
O
 mg 
4
2
2
F  mg
3
4.
Total disc
I
4MR 2
 2MR 2
2
By symmetry I1 
5.
2MR 2 1
 MR 2
4
2
3
I  MR 2
2
3
L2
3  L3
 L 2 
2
4
8 2
6.
Using parallel axes theorem, moment of inertia about ‘O’
2R
2R
2R
2R
2R
2R
I 0  I cm  md 2 


7MR 2
55MR 2
2
 6 M   2R  
2
2
Again, moment of inertia about point P, Ip  I0  md 2
SEC: Jr.Super60_STERLING BT
Page 2
Sri Chaitanya IIT Academy, India

7.
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
2
55MR
181
2
 7M  3R  
MR 2
2
2
  r  F  0.05Nm
    0.05  2  0.031J
8.
1 2
mv  mg  h1  h 2 
2
9.
Conservation of energy
3
mg  4a   mga  mv2com
4
2
4gs  v cm
 vcom  4ga
10.
  R.F  I
2R.T 
2T 
T
MR 2

2
MR
2
Ma
4
Mg  T  ma
mg 
ma
 ma
4
M

mg  a   m 
 4

a
11.
4mg
M  4m
Using principal of conservation of energy
1
1
 m1  m 2  gh   m1  m2  v2  I 2
2
2
  m1  m 2  gh 
1
1
2
 m1  m2 R   I 2  v  R 
2
2
  m1  m 2  gh 
2
 
 m1  m 2  R 2  I 
2 
2  m1  m 2  gh
 m1  m 2  R 2  I
SEC: Jr.Super60_STERLING BT
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Sri Chaitanya IIT Academy, India
12.
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
For Ball
Using parallel axes theorem, for ball moment of inertaia,
2
22
2
I ball  MR 2  M  2R   MR 2
5
5
For two balls Iball
M  2R 
2
MR 2
 2  MR 2 and I rod 

5
12
3
Isystem  I balls  Irod 
44
MR 2 137
MR 2 

MR 2
5
3
15
2
13.
Here vcm   R
1
1 2
1 2
K rot  Icm 2     2  mvcm
2
2 5
5
K trans 
1 2
mvcm
2
1 2
1
mvcm
K tot
2
5


 5 
1
1
7
K total
7
mv2  mv 2
5 cm 2 cm 10
14.
F  f r  ma............(i)
f r R  I 
mR 2
 ............(ii)
2
For pure rolling
a   R............(iii)
From (i)(ii) and (iii)
F
15.
mR
3
2F
 m R  F  mR   
2
2
3mR
Applying energy conservation principle of
 3v 2 
1
1
mv 2  I 2  mg 

2
2
 4g 
1
1 v2 3
 mv2  I 2  mv 2  v  R 
2
2 R
4
SEC: Jr.Super60_STERLING BT
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Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
2
1 v
3
1
1
1
 I 2  mv 2  mv 2  mv 2  I  mR 2
2 R
4
2
4
2
This is the moment of inertia of the disc hence the object is disc.
16.
In pure rolling, the point of contact is the instantaneous centre of rotation of all the
particles of the disc. On applying v  r
We find  is same for all the particles then v  r .
rQ  rC  rp  v Q  v C  v p
17.
Taking torques about reference
x
3w L w
L
  ;x 
4
2 4
6
From end
L L L
 
2 6 3


i
18.
j
  
L  r  P  u cos  t u sin  t  gt 2
u cos 
u cos  gt
Here u  v,  600 , t 
19.

k
0
0
V sin 
g
It is similar to horizontal projection from the top of power of height 2r
Range (R)=velocity  time
 2v
2h
2  2r
 2v
g
g
The distance Ac  4v
SEC: Jr.Super60_STERLING BT
r
g
Page 5
Sri Chaitanya IIT Academy, India
20.
Conceptual
21.
Angular momentum
L  m( r  v )
 1 (3i  j )  (3i  k )
 i  3 j  3k
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
L  1  9  9  19
22.
Let mass of scale be ‘m’. Then,
About ‘P’  net  0
 2 102 g  30  mg  10  0
 6  101  10m  0  0.6  10m
m
23.
0.6
 m  0.06kg
10
Let  be the surface mass density.
Moment of inertia of the lamina about axes passing through ‘O’
2
1
2
2
1

I O     2R     2R     R 2     R 2   R 2 


2
2


13
 R 4
2
Moment of inertia of the lamina about axes passing through ‘P’
2
2 1
I P  8 R 4    2R    2R     R 2  R 2    R 2 
2

 2R 
2


 R2 

 24 R 4  5.5 R 4  18.5 R 4

IP 18.5 R 4 37


3
IO 13  R 4 13
2
SEC: Jr.Super60_STERLING BT
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Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
24.
As per question, same rotational kinetic energy for wheel P and Q
1
1
2
2
 KE rot1  KE rot 2 or , I1 1   I2 2 
2
2
2
 v 
v
I1 
  I2  
3R


R

25.
2
v

   R 
I1 9

I2 1
 
0  r  F 
R
 mg
2
y
x
R
2

mg
u 2 sin 2
1
 mg  mu 2 sin 2
2g
2
1 2
L2
I ; KE  ; E L2 , E I1
2
2I
26.
KE 
27.
From conservation of angular momentum
m 2gH
L mL2
L

 mv
2
12
2
2gH   v 
v 

L
2
[This is the equation of e is applied only at the point of collision]
gH
2
3 2gh
L
2
7
16MR 2
MR 2  2  MR 2 
5
5
5
28.
I
29.
From LOCAM  I11  I22
1100  2.2  I 2  2.I1  ; 2  50rad / s
SEC: Jr.Super60_STERLING BT
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Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
1
1
K.E loss  I112  I222
2
2
1
1
 11002   2  2500
2
2
 5000  2500J  2500J  2.5J
30.
Given, mass of the circular disc, m  20kg
Radius of the circular disc, R  0.2m
Torque required to attain angular speed of 50 rad s 1 ,
  I
 

F.R
  F.R 
I
20  0.2
 10 rad / s 2
1
 20  0.22
2
Using,  2  02  2
  50   0 2  2 10     125rad
2
No.of revolutions, n 
SEC: Jr.Super60_STERLING BT
125
 20
2
Page 8
Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
CHEMISTRY
31.
Conceptual
32.
33.
34.
Inert pair effect
Conceptual
35.
Conceptual
36.
37.
Be
38.
39.
2 Bridge hydrogen
Conceptual
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
Conceptual
60.
sp 3  sp3
BF3
H 3 BO3
Ca 2 B6 O11 5H 2O
44 8
SEC: Jr.Super60_STERLING BT
Page 9
Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
MATHEMATICS
61.
A
B
C
D
 

AB  AC  2AD
 1
 AD 
3i  4k  5i  2j  4k
2

 
  i  j  4k
Length of AD  1  1  16  18
62.
 


Given       a .......(i)




      b .......(ii)
   

From(i),          a  1  .....(iii)
   

From(ii),          b  1  .....(iv)
From(iii) and (iv), we get


 a  1    b  1  .........(v)


Since  is not parallel to  ,
From (v), a  1  0 and b  1  0
   
From (iii),         0
63.
     
apbpcp  0
  
 abc
or p 
3
Hence, P is centroid
64.
Here AB  xi  yj
AC  xi  zk
AD   x  1 i  j  k
As these vectors are coplanar.
x
 x
y
0
0
z  0 
x  1 1 1
SEC: Jr.Super60_STERLING BT
1 1 1
  1
x y z
Page 10
Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
65.
66.
5c  13b  c 
c
13
b
5
13
b 2b
5
67.
AB is along the x-axis and BD is along the y-axis.

AB  2i  AB  BC  CD  ......  2
From the figure, BM  BC sin 600
 2 sin 600  3

 BD  2 3 j

 

BC  BC cos 600 i  BC sin 60 0 j  i  3 j
 



 CD  BD  BC  2 3 j  i  3 j  i  3 j

68.

Let the given position vectors be of points A,B and C, respectively. Then

AB 
               
2

BC 
              
2

CA 
               
2
2
2
2
2
2
2
  
 AB  BC  CA
Hence, ABC is an equilateral triangle
SEC: Jr.Super60_STERLING BT
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Sri Chaitanya IIT Academy, India

69.
2
1
1

1
1
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
1
1  0   2   2 1   2   0
2
 2
  2
Hence, two real solutions.
70.


a  i  k and b  xi  j  1  x  k

c  yi  xj  1  x  y  k
1
1 0
 
abc   x 1


1 x
y x 1 x  y
 1  x  y  x  x 2   1 x 2  y   1
71.
 
  

a  2b  1 c, b  3c  2 a

 

a  2 2 a  3c  1 c





a 1  22  a   1  6  c ,so 1  6
  
a  2b  6c  0
72.
Conceptual
73.
   
 OB  OD OA  OC
OG 

2
2
   
 

OA  OB  OC  OD  2 OA  OC  4OG

74.


n  cos  cos  i  cos  sin  j  sin  k

n  cos 2  cos 2  i  cos 2  sin 2  j  sin 2  k 
  cos 2   sin 2    1
75.
Conceptual
SEC: Jr.Super60_STERLING BT
Page 12
Sri Chaitanya IIT Academy, India
76.
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s



Let p  ai  aj  ck, q  i  k and r  ci  cj  bk
a a c
 
 pqr   1 0 1  0
 
c c b
a  0  c  a  b  c  c c  0
ac  ab  ac  c 2  0
c 2  ab
77.
1
5
f    i  j  2k
4
4
 i   j k   
78.
1
,  2
4
The bisector divides BC in the ratio AB:AC
i.e 6 : 3 6 or1: 3 at pint D.
The position vector of D is
Hence, AD 
1 5
3 9
i  j  3k and AD  i  j  0k
4 4
4 4
9 81 3


10
16 16 4
79.
Conceptual
80.
Vector a bisects the angle between vectors b and c .




 i  2j  k 
 a   b  c   i  2j   k   


2 



  2 and   2 and   2 
   1and   1
81.
    
2
9PG 2  AB AB  BC
2
    PB   PC 
 3  PA

2
2
 9  4   c2  b2  a 2  31  9  4
82.
a 2  b 2  c 2  42  36  6

Let R be the resultant. Then
   
R  F1  F2  F3   p  1 i  4j  p  2, 4
SEC: Jr.Super60_STERLING BT
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Sri Chaitanya IIT Academy, India
83.
OP 
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
OC  4OB
3OA  2OC
, OQ 
5
5
 OR 
3OB  7OA
OB  3OA
, OS 
10
4
AP  BQ  CR  OP  OA  OQ  OB  OR  OC

84.
3OA  OB  4OC 3OS  4OC 2

 CS
10
10
5
     
 
L.H.S.  d  a  d  b  h  c  3 g  h


AP  BQ  CR 2
AP  BQ  CR
2
 

5
5
CS
CS

  
abc
   

 2d  a  b  c  3
 2h
3


 

 2d  2h  2 d  h  2HD    2


85.





 
1
ai  bj  k   i  2j  m j  2k
2
 a  , b  2  m and m 

1
4
1
 ai  bj  k is unit vector
2
 a 2  b2 
3
11
 5a 2  a   0
4
16
a1 and a 2 are roots of above equation

86.
1 1 a1  a 2
16
 

a1 a 2
a 1a 2
11
Clearly the vector along the longer diagonal is
 
a  b  3i  6j  2k .
 
Hence, the length of the longer diagonals is a  b  3i  6j  2k  7
87.
 

Let P.V.of A,B and D be 0, b and d , respectively.
SEC: Jr.Super60_STERLING BT
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Sri Chaitanya IIT Academy, India
02-09-23_Jr.Super60_STERLING BT_Jee-Main_WTM-12_Key & Sol’s
  
Then P.V. of C, c  b  d


 d
 b
Also P.V.of A1  b  and P.V.of B1  d 
2
2
  3  
3 
 AA1  AB1  b  d  AC
2
2
 
 
 


 
   1 a1  a 2   a 2  a 3   a 3  a 4  2a 2  a 3   a 4  0

88.


 

 





i.e.,    1 a1  1      2  a 2       1 a 3      a 4  0
  

Since, a1 , a 2 , a 3 and a 4 are linearly independent,
  1  0,1      2  0,     1  0,     0
2
3
1
3
i.e.,   1,    ,    ,  
89.
1
3
Let AB  b, AC  c, AD  d P divides DC in the ratio 1: 2  AP 

c
2
Q is the midpoint of AC  AQ  

 kPQ  AB  2AD  BC  2DC

 
 b  2d  c  b  2 c  d
1
90.
c  2d
3
c  4d
6

1343 1
1
1
1  0  2  3040  0    2015
1

2
SEC: Jr.Super60_STERLING BT
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